1
$\begingroup$

I tried doing these questions and just wanted to know if there are any that I should look back at and fix. Please don't fix anything this looks exactly how it does on my paper and in my book.

For the last question I am going to try to explain this the best that I can. I wanted to make it look like a "-----|----|" for calculating the answer.

I wanted the 0.0030 moles of $\ce{HCl(aq)}$ to start out at the top left of that structure I typed out. Then have the 1 mole $\ce{Na2CO3(s)}$ on the top next line over and then have the 2 moles $\ce{HCl(aq)}$ below it. Then when calculated it would equal 0.0015 moles of $\ce{Na2CO3(s)}$.

And

I wanted the 0.0015 moles of $\ce{Na2CO3(s)}$ to start out at the top left of that structure I typed out. Then have the 105.99g/mol $\ce{Na2CO3(s)}$ on the top next line over and then have the 1mol $\ce{Na2CO3(s)}$ below it. Then when calculated it would equal 0.16g $\ce{Na2CO3(s)}$.

If you still don't understand I am going to attempt to type the structure out with the equations.

Please use this link if you want a better look at the equations and parts.

https://books.google.com/books?id=X3KX6p8hnlIC&pg=PA304&lpg=PA304&dq=Five+repeated+calculations+are+required+to+establish+the+data+plot+for+the+determination+of&source=bl&ots=AenLh4IGsI&sig=hs7pdP6AnhvgaTVs0zvcGChbTzE&hl=en&sa=X&ved=0ahUKEwjzpZ_WlKrJAhWBHD4KHQ-ICqIQ6AEIHDAA#v=onepage&q=Preparation%20of%20Borax%20solutions&f=false

For "2. a. Experimental Procedure, Part A.2." in the link it is on page 303 the last 3 is cut out on the page for some reason and it is called "2. Prepare stock solution of borax.". For some reason the book shows it as Part B.2, some things are moved around a bit on the lab book compared to mine.

For "2. b. Experimental Procedure, Part A.3." in the link it is on page 303 the last 3 is cut out on the page for some reason and it is called "3. Prepare the test solutions of borax.". For some reason the book shows it as Part B.3, some things are moved around a bit on the lab book compared to mine.

  1. a. Experimental Procedure, Part A.2. Describe the preparation of 200 mL of 0.20 M HCl starting with conc HCl (12.1M).

M1V1 = M2V2

M1 = 12.1M

M2 = 0.20M

V2 = 0.20L

(12.1M)(V1) = (0.20M)(0.20L)

(0.20L X 0.20M) / 12.1M = 0.0033L

V1 = 0.0033L

The description for the preparation of 200mL of 0.20M HCl starting with conc HCl (12.1M) is adding 0.0033L (V1) of the solvent, water, with concentrated HCl (12.1M) to get 0.20L (V2) of 0.20M (M2) HCl into a volumetric flask and make sure to stop filling up at the etched line and to mix these solutions.

b. Experimental Procedure, Part A.3. How many grams of the primary standard sodium carbonate, $\ce{Na2CO3}$ (molar mass = 105.99g/mol) is needed to react with 15mL of 0.20 M HCl?

moles of HCl(aq) = MHCl(aq)•LHCl(aq)

0.20M HCl(aq) • 0.015L HCl(aq) (15mL = 0.015L)

=0.0030 moles of HCl(aq)

$\ce{Na2CO3(s)}$ + 2 HCl(aq) → $\ce{H2O(l)}$ + $\ce{CO2(g)}$ + 2 $\ce{NaCl(aq)}$

The reactants of the chemical equation above have one mole of $\ce{Na2CO3(s)}$ react with two moles of HCl(aq).

     0.0030 moles of HCl(aq)            |      1 mole Na2CO3(s)
                                               2 moles HCl(aq)

= 0.0015 moles of $\ce{Na2CO3(s)}$

      0.0015 moles of Na2CO3(s)         |      105.99 g/mol Na2CO3(s)
                                               1 mol Na2CO3(s)

= 0.16g $\ce{Na2CO3(s)}$

The number of grams of the primary standard sodium carbonate, $\ce{Na2CO3(s)}$ (molar mass = 105.99g/mol) is needed to react with 15mL of 0.20M HCl is 0.16g $\ce{Na2CO3(s)}$

$\endgroup$
1
$\begingroup$

As @Marzipanherz said there is a miscalculation in part A. The 200 ml in brackets at the end should be written in L ( 0.2L ).

I got v1 = 0.0033 L using the same equation.

Also add what state the material is in (eg. g, l, s, aq) when writing the chemical equations.

$\endgroup$
  • $\begingroup$ Are you talking about in part b? $\endgroup$ – Josh Palmer Nov 25 '15 at 13:37
  • $\begingroup$ I am going to edit it and please tell me that what I am editing is what you were talking about. $\endgroup$ – Josh Palmer Nov 25 '15 at 13:40
  • $\begingroup$ Part A, 0.2 X 0.2/ 12.1 = 0.0033. Part B is ok. $\endgroup$ – Technetium Nov 25 '15 at 14:04
  • $\begingroup$ I fixed the part about Part A on here. Is there anything else? $\endgroup$ – Josh Palmer Nov 25 '15 at 14:15
  • $\begingroup$ I need to ask you a question. It's about a different question than the one stated here. $\endgroup$ – Josh Palmer Nov 25 '15 at 18:02
0
$\begingroup$

In part a seems to be a calculation issue in the last step. And I would expect the name of the second liquid when you describe the mixing.

Part b looks good to me.

$\endgroup$
  • $\begingroup$ I fixed something, but I don't know if it was what you were asking about please tell me if I fixed it or not. $\endgroup$ – Josh Palmer Nov 25 '15 at 13:49
  • $\begingroup$ The formulas and results seem to be ok now. However, your written description makes me wonder whether you understood the preparation. Take deep breath. Imagine, you were in a lab. Read the task carefully and re-think: which liquids will you mix? $\endgroup$ – Arsak Nov 25 '15 at 14:01
  • $\begingroup$ How does it look now? $\endgroup$ – Josh Palmer Nov 25 '15 at 14:15
  • $\begingroup$ Solving the equation for V1 is not correct now, although you have put the right result at the end. Please have a look at @Joel 's comment. And your description is still wrong: you can't use 200ml of the 0,2M HCl, because this is what you want to produce! $\endgroup$ – Arsak Nov 25 '15 at 16:02
  • $\begingroup$ How is the equation for V1 incorrect now? I just don't know what part is wrong now. I saw the way he got the 0.0033L and I thought I put it into a similar way. I just wanted to know what was wrong there. Also, I will try to fix the description again. $\endgroup$ – Josh Palmer Nov 25 '15 at 16:14

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.